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# Find q when 22.0 g of water is heated from 45.0°C to 100.0°C.?

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- jacob sLv 74 weeks agoFavorite Answer
We know the relationship

q = m*s*∆T

Where q is heat, m is mass of substance, s is specific heat

∆T is change in temperature

From Question,

m = 22.0 g

specific heat of water, s = 4.184 Jg-1oC-1

∆T = (100-45)oC = 55oC

q = 22.0 g *4.184 Jg-1oC-1 *55oC

q = 5062.64 J

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